Julia中基准和时间宏之间的区别 [英] Difference between benchmark and time macro in Julia

问题描述

我最近发现了两个宏之间的巨大差异:@benchmark 和@time 在内存分配信息和时间方面.例如:

I've recently discovered a huge difference between two macros: @benchmark and @time in terms of memory allocation information and time. For example:

@benchmark quadgk(x -> x, 0., 1.)
BenchmarkTools.Trial: 
memory estimate:  560 bytes
allocs estimate:  17
--------------
minimum time:     575.890 ns (0.00% GC)
median time:      595.049 ns (0.00% GC)
mean time:        787.248 ns (22.15% GC)
maximum time:     41.578 μs (97.60% GC)
--------------
samples:          10000
evals/sample:     182

@time quadgk(x -> x, 0., 1.)
0.234635 seconds (175.02 k allocations: 9.000 MiB)
(0.5, 0.0)

为什么这两个例子有很大的不同?

Why there is a big difference between these two examples?

推荐答案

原因是预编译开销.要查看此定义:

The reason is precompilation overhead. To see this define:

julia> h() = quadgk(x -> x, 0., 1.)
h (generic function with 1 method)

julia> @time h()
  1.151921 seconds (915.60 k allocations: 48.166 MiB, 1.64% gc time)
(0.5, 0.0)

julia> @time h()
  0.000013 seconds (21 allocations: 720 bytes)
(0.5, 0.0)

相对于

julia> @time quadgk(x -> x, 0., 1.)
  0.312454 seconds (217.94 k allocations: 11.158 MiB, 2.37% gc time)
(0.5, 0.0)

julia> @time quadgk(x -> x, 0., 1.)
  0.279686 seconds (180.17 k allocations: 9.234 MiB)
(0.5, 0.0)

这里发生的情况是，在第一次调用中，将 quadgk 包装在一个函数中，匿名函数 x->x 只定义一次，因为它被包装在一个函数，因此 quadgk 只编译一次.在第二次调用中，每次调用都会重新定义 x->x，因此每次都必须执行编译.

What happens here is that in the first call, wrapping quadgk in a function, anonymous function x->x is defined only once, since it is wrapped in a function, and thus quadgk is compiled only once. In the second call x->x is defined anew with every call and thus compilation has to be performed each time.

现在关键是 BenchmarkTools.jl 将您的代码包装在一个函数中，您可以通过检查 generate_benchmark_definition 函数在该包中的工作方式来检查该函数，因此它等效于上面介绍的第一种方法.

And now the crucial point is that BenchmarkTools.jl wraps your code in a function which you can check by inspecting how generate_benchmark_definition function works in this package, so it is equivalent to the first approach presented above.

在不重新定义优化函数的情况下运行代码的另一种方法是:

Another way to run the code without redefining the optimized function would be:

julia> g(x) = x
g (generic function with 1 method)

julia> @time quadgk(g, 0., 1.)
  1.184723 seconds (951.18 k allocations: 49.977 MiB, 1.58% gc time)
(0.5, 0.0)

julia> @time quadgk(g, 0., 1.)
  0.000020 seconds (23 allocations: 752 bytes)
(0.5, 0.0)

(虽然这不是 BenchmarkTools.jl 所做的——我添加它是为了表明当您使用函数 g 时，您无需支付两次预编译税)

(though this is not what BenchmarkTools.jl does - I add it to show that when you use function g you do not pay precompilation tax twice)

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